Calculate the relative lowering of vapour pressure of a solution containing $46 \ g$ of non-volatile solute in $162 \ g$ of water at $20^{\circ} C$. [Molar mass of non-volatile solute $= 46 \ g \ mol^{-1}$]

When a certain amount of solid $A$ is dissolved in $100 \ g$ of water at $25^{\circ} \ C$ to make a dilute solution,the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76 \ mm \ Hg$. The number of moles of solute $A$ added is $...........$ (Nearest Integer).

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water,what is the vapour pressure (in $mm$) of the resultant solution?

Vapour pressure of a solution is

The vapour pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water (in $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$